Chapter 11, Problem 11C.4P

Interpretation Introduction

Interpretation:

The validation of the statement that the vibrational levels of ${}^{23}{\text{Na}}^{127}\text{I}$ that lie at the given wavenumbers fit the expression $\left(\nu +\frac{1}{2}\right)\tilde{\nu }-{\left(\nu +\frac{1}{2}\right)}^2{x}_e\tilde{\nu }$ has to be shown.  The value of force constant, zero point energy and dissociation energy of the molecule is to be calculated.

Concept introduction:

A molecule is made up of atoms that are bonded together by covalent bonds.  These bonds undergo a to and fro movement to vibrate.  The vibrational frequency of a bond is given by the expression shown below.

    $v=\frac{1}{2\text{π}}\sqrt{\frac{k_f}{\mu }}$

Answer to Problem 11C.4P

The statement that the vibrational levels of ${}^{23}{\text{Na}}^{127}\text{I}$ that lie at the given wavenumbers fit the expression $\left(\nu +\frac{1}{2}\right)\tilde{\nu }-{\left(\nu +\frac{1}{2}\right)}^2{x}_e\tilde{\nu }$ has been shown.  The value of zeta potential is $\underset{\_}{142.8c{m}^{-1}}$.  The value of force constant is $\underset{\_}{142.8c{m}^{-1}}$.  The value of dissociation energy is $\underset{\_}{324kJmo{l}^{-1}}$.

Explanation of Solution

The expression for vibrational energy levels is shown below.

    $\tilde{\nu }=\left(\nu +\frac{1}{2}\right){\tilde{\nu }}_0-{\left(\nu +\frac{1}{2}\right)}^2{x}_e{\tilde{\nu }}_0$

Where,

• $\nu$ is the vibrational quantum number.
• $\tilde{\nu }$ is the wavenumber.
• ${\tilde{\nu }}_0$ is the harmonic wavenumber of the oscillator.
• $x_{\text{e}}$ is the anharmonicity constant.

The difference between two vibrational energy levels is expressed as shown below.

    $\Delta \tilde{\nu }=\tilde{\nu }\left(\nu +1\right)-\tilde{\nu }\left(\nu \right)$

The above expression can be expanded as shown below.

    $\begin{array}{c}\tilde{\nu }\left(\nu +1\right)-\tilde{\nu }\left(\nu \right)=\left(\nu +1+\frac{1}{2}\right){\tilde{\nu }}_0-{\left(\nu +1+\frac{1}{2}\right)}^2{x}_e{\tilde{\nu }}_0-\left(\nu +\frac{1}{2}\right){\tilde{\nu }}_0+{\left(\nu +\frac{1}{2}\right)}^2{x}_e{\tilde{\nu }}_0\\ =\left(\nu +\frac{3}{2}\right){\tilde{\nu }}_0-{\left(\nu +\frac{3}{2}\right)}^2{x}_e{\tilde{\nu }}_0-\left(\nu +\frac{1}{2}\right){\tilde{\nu }}_0+{\left(\nu +\frac{1}{2}\right)}^2{x}_e{\tilde{\nu }}_0\\ =\left(\nu +\frac{3}{2}\right){\tilde{\nu }}_0-\left({\nu }^2+3\nu +\frac{9}{4}\right)x_e{\tilde{\nu }}_0-\left(\nu +\frac{1}{2}\right){\tilde{\nu }}_0+\left({\nu }^2+\nu +\frac{1}{4}\right)x_e{\tilde{\nu }}_0\\ =\left(\nu +\frac{3}{2}-\nu -\frac{1}{2}\right){\tilde{\nu }}_0-\left({\nu }^2+3\nu +\frac{9}{4}-{\nu }^2-\nu -\frac{1}{4}\right)x_e{\tilde{\nu }}_0\end{array}$

The above expression can be further simplified as shown below.

    $\begin{array}{c}\Delta \tilde{\nu }=\left(\frac{2}{2}\right){\tilde{\nu }}_0-\left(2\nu +\frac{8}{4}\right)x_e{\tilde{\nu }}_0\\ \Delta \tilde{\nu }={\tilde{\nu }}_0-2\left(\nu +1\right)x_e{\tilde{\nu }}_0\end{array}$        (1)

The above expression is to be compared with the equation for the straight line, that is, $y=mx+c$.

Where,

• $y$ is the component on $y$ axis.
• $x$ is the component on $x$ axis.
• $m$ is the slope.
• $c$ is the intercept.

By comparing the equations, a graph can be plotted by taking $\Delta \tilde{\nu }$ on the $y$ axis and $\left(\nu +1\right)$ on the $x$ axis.  The slope of the graph will be $-2x_e{\tilde{\nu }}_0$ and the value of intercept will be ${\tilde{\nu }}_0$.  The graph is plotted as shown below.

Figure 1

From the graph in Figure 1, the value of intercept, that is, ${\tilde{\nu }}_0$ is equal to $286{\text{cm}}^{-1}$.

The value of slope is calculated as shown below.

    $\begin{array}{c}\text{Slope}=\frac{283-284.5}{2-1}\\ =-1.5\end{array}$

Therefore, the value of slope, that is $-2x_e{\tilde{\nu }}_0$ is equal to $-1.5{\text{cm}}^{-1}$.

The value of $x_e$ is calculated as shown below.

    $\begin{array}{c}-2x_e{\tilde{\nu }}_0=-1.5\\ x_e=\frac{-1.5{\text{cm}}^{-1}}{-2\times {\tilde{\nu }}_0}\\ =\frac{-1.5{\text{cm}}^{-1}}{-2\times 286{\text{cm}}^{-1}}\\ =0.002622\end{array}$

It can be seen that there is a high value of linearity in the graph.  Therefore, the given data fits well with the expression.

The value of zeta potential can be calculated as shown below.

    $\begin{array}{c}E\left(\nu =0\right)=\left(0+\frac{1}{2}\right){\tilde{\nu }}_0-{\left(0+\frac{1}{2}\right)}^2{x}_e{\tilde{\nu }}_0\\ =\left(\frac{1}{2}-\frac{1}{4}{x}_e\right){\tilde{\nu }}_0\\ =\left(\frac{1}{2}-\frac{1}{4}\left(0.002622\right)\right)\times 286{\text{cm}}^{-1}\\ =\underset{\_}{142.8c{m}^{-1}}\end{array}$

Therefore the value of zeta potential is $\underset{\_}{142.8c{m}^{-1}}$.

The mass of sodium atom is $23\text{u}$.

The mass of sodium atom in $\text{kg}$ is converted as shown below.

    $\begin{array}{c}{m}_{\text{Na}}=\left(23\text{u}\right)\left(\frac{1.66054\times {10}^{-27}\text{kg}}{1\text{u}}\right)\\ =3.8192\times {10}^{-26}\text{kg}\end{array}$

The mass of iodine atom is $127\text{u}$.

The mass of iodine atom in $\text{kg}$ is converted as shown below.

    $\begin{array}{c}{m}_{\text{I}}=\left(127\text{u}\right)\left(\frac{1.66054\times {10}^{-27}\text{kg}}{1\text{u}}\right)\\ =2.1089\times {10}^{-25}\text{kg}\end{array}$

The reduced mass of a molecule $\left(\mu \right)$ is given by the expression as shown below.

  $\mu =\frac{m_{\text{A}}{m}_{\text{B}}}{m_{\text{A}}+m_{\text{B}}}$        (2)

Where,

• $m_{\text{A}}$ is the mass of atom $\text{A}$.
• $m_{\text{B}}$ is the mass of atom $\text{B}$.

Substitute the values of the mass of sodium atom and iodine atom in the equation (2).

  $\begin{array}{c}{\mu }_{\text{NaI}}=\frac{\left(3.8192\times {10}^{-26}\text{kg}\right)\left(2.1089\times {10}^{-25}\text{kg}\right)}{\left(3.8192\times {10}^{-26}\text{kg}\right)+\left(2.1089\times {10}^{-25}\text{kg}\right)}\\ =3.2336\times {10}^{-26}\text{kg}\end{array}$

The wavenumber $\left(\overline{v}\right)$ for the corresponding vibrational frequency of a bond is given by the expression shown below.

    $\overline{v}=\frac{1}{2\text{πc}}\sqrt{\frac{k_f}{\mu }}$

Where,

• $k_f$ is the force constant of the bond.
• $\mu$ is the reduced mass of the molecule.
• $c$ isthe speed of light with value of $2.998\times {10}^8\text{m}{\text{s}}^{-1}$.

Rearrange the above equation for the value of $k_f$.

    $k_f=\mu {\left(2\text{πc}\overline{v}\right)}^2$        (3)

Substitute the value of wavenumber of the corresponding vibrational frequency of $\text{NaI}$ bond and reduced mass of $\text{NaI}$ in the equation (3).

    $\begin{array}{c}{k}_f\left(\text{NaI}\right)=\left(3.2336\times {10}^{-26}\text{kg}\right){\left(2\text{π}\left(2.998\times {10}^8\text{m}{\text{s}}^{-1}\right)\left(286\times {10}^2{\text{m}}^{-1}\right)\right)}^2\\ =93.8\text{kg}{\text{s}}^{-2}\\ =\underset{\_}{93.8N{m}^{-1}}\end{array}$

Therefore, the value of force constant is $\underset{\_}{142.8c{m}^{-1}}$.

The value of anharmonicity coefficient is calculated as shown below.

    $x_e=\frac{h{\nu }_0}{4D_e}$        (4)

Rearrange the above equation to obtain the value of $D_e$.

    $\begin{array}{l}{x}_e=\frac{h{\nu }_0}{4D_e}\\ D_e=\frac{h{\nu }_0}{4x_e}\end{array}$        (5)

The value of dissociation energy is calculated as shown below.

    $D_0=D_e-\text{Zeta}\text{potential}$        (6)

Substitute the value of equation (5) in equation (6).

    $\begin{array}{c}{D}_0=D_e-\text{Zeta}\text{potential}\\ \text{=}\frac{h{\nu }_0}{4x_e}-\text{Zeta}\text{potential}\\ \text{=}\left(\begin{array}{c}\frac{\left(6.626\times {10}^{-34}\text{J}\text{s}\right)\left(2.998\times {10}^8\text{m}{\text{s}}^{-1}\right)\left(286\times {10}^2{\text{m}}^{-1}\right)}{4\times 0.002622}-\\ \left(6.626\times {10}^{-34}\text{J}\text{s}\right)\left(2.998\times {10}^8\text{m}{\text{s}}^{-1}\right)\left(142.8\times {10}^2{\text{cm}}^{-1}\right)\end{array}\right)\\ =\underset{\_}{324kJmo{l}^{-1}}\end{array}$

Therefore, the value of dissociation energy is $\underset{\_}{324kJmo{l}^{-1}}$.